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0

If it is an isometry then it has a left inverse that is: $$\langle x,LRz\rangle=\langle x,z\rangle=\langle Rx,Rz\rangle$$ So the adjoint must act as a left inverse too. On the other hand if the adjoint acts as a left inverse then: $$\langle Rx,Ry\rangle=\langle x,R^*Ry\rangle=\langle x,y\rangle$$ Thus it is an isometry. (Note, there's no need of ...


0

If it is an isometry then it has a left inverse that is: $$\langle x,LRz\rangle=\langle x,z\rangle=\langle Rx,Rz\rangle$$ So the adjoint must act as a left inverse too. On the other hand if the adjoint acts as a left inverse then: $$\langle Rx,Ry\rangle=\langle x,R^*Ry\rangle=\langle x,y\rangle$$ Thus it is an isometry. (Note, there's no need of ...


0

Notice that the range of the unit ball is contained in the compact set $$K:=\left\{\sum_{j=1}^na_j\phi_j, |a_j|\leqslant \lVert \psi_j\rVert_2\right\}.$$


1

In fact if $T$ is non-expansive then the proposed inequality follow from the Cauchy-Schwartz inequality, since with notation of the proposer we have $$ \langle T(x)-T(y),x-y\rangle= \langle A(x-y),x-y\rangle\leq\Vert A(x-y)\Vert\,\vert x-y\Vert\leq\Vert x-y\Vert^2. $$ Now, the converse is not correct in general without supplementary assumptions, for ...


5

$|\langle x,y\rangle|\le\|x\|\|y\|=(\sqrt{2\epsilon}\|x\|)(\frac{\|y\|}{\sqrt{2\epsilon}})\leq\frac{1}{2}(2\epsilon\|x\|^2+\frac{\|y\|^2}{2\epsilon})=\epsilon\|x\|^2+C_{\epsilon}\|y\|^2$ The last inequality is a variant of the geometric inequality $ab\leq\frac{1}{2}(a^2+b^2)$.


0

Given an arbitrary (infinite-dimensional) Hilbert space $\mathcal H$, you can find an orthonormal sequence $x_n$ in it (inductively, by choosing a unit vector $x_n$ from $\{x_1,\dots,x_{n-1}\}^\perp$). Let $M$ be the closed linear span of $\{x_n:n=1,2,\dots\}^\perp$, and $N=M^\perp$. On the space $M$ you have the natural analogues of $T_l$ and $T_r$. They ...


1

Let $H$ be a complex Hilbert space and let $T \in B(H)$ be an isometry. We claim that $T^\ast T = I$. To this end let $h \in H$ and note that by the Hilbert space Riesz representation theorem a linear functional in $H^\ast$ corresponds to an element $h \in H$ ($h \mapsto \langle \cdot, h \rangle$). Also note that if $\varphi (x) = 0$ for all $\varphi \in ...


0

You're right that Cauchy-Schwarz is important. It comes in here: $$\langle x, Ax \rangle \leq ||x||_1 ||Ax||_1$$ Can you fill in the rest?


1

I think the author is being a little sloppy in this paragraph, which raised your question: Lower semicontinuity was earlier defined by the openness of upper level sets. The weak topology being non-metrizable, one should not casually insert "in other words" followed by the definition of sequential weak lower semicontinuity. For example, Ciarlet ...


1

$T$ must be densely defined for $T^\ast$ to be defined. That $T^\ast$ is a left inverse of $T$ implies that $\mathcal{R}(T) \subset \mathcal{D}(T^\ast)$. And hence for every $x \in \mathcal{D}(T)$ we have $$\langle x,x\rangle = \langle x, T^\ast T x\rangle = \langle Tx,Tx\rangle,$$ or in other words $\lVert Tx\rVert = \lVert x\rVert$, i.e. $T$ is an ...


1

Boundedness of $X$ is needed just to conclude that $\phi$ is finite. $\|h-x\| \le \|h'-x\| + \|h-h'\|$. Hence $\|h-x\| \le \phi(h') + \|h-h'\|$, and so $\phi(h) \le \phi(h') + \|h-h'\|$. Reversing the roles of $h,h'$ gives $|\phi(h)- \phi(h')| \le \|h-h'\|$.


3

One can prove strong convergence following this article: The product of projection operators I. Halperin http://acta.fyx.hu/acta/showCustomerArticle.action?id=7164&dataObjectType=article The original result is due to Neumann. Halperin's note generalizes this to $m$ projections: There it is proven that $(P_1P_2\dots P_m)^n x \to z$ with ...


1

Hint: consider a piecewise function $f_n$, taking the value $n$ in $0$, affine on $(0,1/n^2)$, and taking the value $0$ from $1/n^2$ to $1$. $\|f_n\|_1 = 1/2n, \|f\|_{[0,1]} = n$.


2

It's ok. Letting $\xi\in X$ denote the integration dummy variable we have $$\left\lvert \int_X \langle x_n(\xi)-x(\xi), h\rangle\, d\xi\right\rvert\le \lVert h\rVert_H\int_X\lVert x_n(\xi)-x(\xi)\rVert_H \, d\xi = \lVert x_n-x\rVert_{L^1(X;H)}\lVert h\rVert_H\to 0.$$


1

The idea is to divide the inequality $$ |\zeta|^2 \|z\|^2 + 2\Re \langle y ,\zeta z\rangle \ge0 $$ by $\zeta$ for suitable values of $\zeta$. First, take $\zeta\in \mathbb R$, $\zeta> 0$, then divide by $\zeta$, which gives $$ |\zeta| \|z\|^2 + 2\Re \langle y , z\rangle \ge0, $$ then $\zeta\searrow 0$. Do the same for a negative $\zeta$. At the end, ...


0

It is enough to show that $\alpha_{T}=\inf \{ (Tx,x) : \|x\|=1\}$ and $\beta_{T}=\sup\{(Tx,x) :\|x\|=1\}$ are in the spectrum of $T$. Suppose $A \in \mathcal{B}(H)$ is selfadjoint with $(Ax,x) \ge 0$ for all $x \in H$. Equivalently $(Ax,x) \ge 0$ for all unit vectors $x$. Then $(x,y)_{A}$ defines a pseudo inner-product (i.e., is positive but maybe not ...


1

It is like you are saying. If you write $P_V$ for the orthogonal projection onto $V$, then $$ \min\{\|x-x_0\|:\ x\in V\}=\|P_Vx_0-x_0\|=\|(I-P_V)x_0\|=\|P_{V^\perp}x_0\|=\max\{|\langle P_{V^\perp}x_0,y\rangle|:\ \|y\|=1\}=\max\{|\langle x_0,P_{V^\perp}y\rangle|:\ \|y\|=1\}=\max\{|\langle x_0,P_{V^\perp}y\rangle|:\ \|y\|=1\} =\max\{|\langle x_0,y\rangle|:\ ...


1

For selfadjoint operators we know that $$ \inf\sigma(T)=\inf\{\langle Tx,x\rangle:x\in S_H\}\\ \sup\sigma(T)=\sup\{\langle Tx,x\rangle:x\in S_H\} $$ It is remains to apply result of this answer.


0

Assume that $x$ is orthogonal to each $f_p$. The function $$h(z):=\sum_{k=0}^\infty x_kz^k$$ is holomorphic on the unit disk and vanishes on the non-discrete set $\{b^p,p\geqslant 1\}$. We conclude that $h$ vanishes identically hence $x=0$.


1

This is correct with the right interpretation of the symbol $\oplus$. On the left it should denote the disjoint union, on the right the Hilbert space sum. You should also be more explicit about the measures you are using (perhaps the spaces are Riemannian manifolds with their natural measures).


0

The Div-Curl lemma gives that $\int u_n v_n \rightarrow \int uv$ if each sequence converges weakly, and the divergence of one sequence and the curl of the other are precompact in $H^{-1}$.


1

What you have proved is $$ \langle \psi_n | U^\dagger U -I | \psi_m\rangle = \delta_{m,n}. $$ Since the $\psi_n$ are a basis this shows $U^\dagger U -I=0$, or equivalently $U^\dagger U =I$.


1

There are certainly instances where this is false. As Jochen said, if this is an orthonormal basis, then $span(\beta-\beta_1) \subset \{\beta_1\}^\perp$ and so is not dense. There are bases where it is possible to remove one or any finite number of terms and keep density, these are called over complete bases. For instance in a RKHS, the span of the kernel ...


1

If $F_N$ converges wekly then it is a bounded sequence in $H$ (this is true in all locally convex spaces). Moreover, by Pythagoras, $\|F_N-F_M\|^2 = \sum\limits_{n=M+1}^N \|f_n\|^2 \to 0$ (for $N>M\to\infty$). Therefore, $F_N$ is a Cauchy sequence and hence convergent.


1

If, for example, $\beta_n$ are orthogonal then span$\lbrace \beta_2,\beta_3,\ldots\rbrace$ is not dense. The distance of $\beta_1$ to this set is $\|\beta_1\|$.


-3

Generally, to prove that $A=B$ for sets $A,B$ one shows that if $a \in A$ then also $a \in B$ (i.e. $A \subseteq B$) and also if $b \in B$ then also $b \in A$ (so $B \subseteq A$). Let $A = L^2(0,\pi)$ and let $B = \mathrm{span}\left(\left\{\sin kx\right\}_{k=0}^\infty\right)$. How do you proceed to apply this to your problem?


1

Hint: For a projection $P$, we have $$\ker P = \operatorname{im} (I - P).$$


3

Since these are two vectors in a same natural basis (consisting of all $\varphi_i\otimes\varphi_j$) for the tensor product space, they are obviously distinct.


0

Suppose that the inequality doesn't hold, so for all $n\in\mathbb{N}$ there exists a $u_n\in H$ such that $$ \|u_n\|_{H^1}>n\sum_{|\alpha|=2}\|D^\alpha u_n\|_{L^2}. $$ Further, normalizing, we can assume that $\|u_n\|_{H^1}=1$ for all $n\in\mathbb{N}$. So, $$ \frac{1}{n}>\sum_{|\alpha|=2}\|D^\alpha u_n\|_{L^2} $$ for all $n\in\mathbb{N}$. You should be ...


3

Suppose that $f=\{ f_{0},f_{1},f_{2},\cdots\} \in l^{2}$ is orthogonal to every such $f_{p}$. Let $$ F(z)=\sum_{n=0}^{\infty}f_{n}z^{n},\;\;\; |z| < 1. $$ $F$ converges absolutely in $|z| < 1$ because the Cauchy-Schwartz inequality implies $$ \left[\sum_{n=0}^{\infty}|f_{n}||z|^{n}\right]^{2} \le ...


0

If $f$ and $g$ are polynomials in $z$, $\overline{z}$, then the result is fairly clear. See if you can generalize from there. You're going to have to be precise about what you mean by $f\circ g$ for Borel functions.


3

Like this: $$\langle (x,y),(z,w)\rangle_{H_1\oplus H_2}=\langle x,z\rangle_{H_1}+\langle y,w\rangle_{H_2}$$


0

If you have a vector space $X$ with an inner product $\langle \cdot, \cdot \rangle$, this defines a norm $\|\cdot\|$ by $\|x\|=\sqrt{\langle x, x\rangle}$ (it is a good exercise to prove that this is in fact a norm). Similarly, this defines a metric, $d(x,y)=\|x-y\|$ (it is again a good exercise to prove that this is in fact a metric). This is the case for ...


1

A Hilbert space is an inner product space that is complete with respect to the norm. Completeness is what differentiates the two. Not every metric space can be defined by an inner product, for instance the space of continuous functions on $[0,1]$ with the supremum norm as its metric doesn't come from an inner product. In particular, a Banach space (a ...


1

I misinterpreted your question. I thought you wanted to see the standard construction of the $V$ where $V^{\star}V=P$ is the orthogonal projection onto $\mathcal{R}(|T|)^{c}$. This $V$ is unique in all cases, even where $\mathcal{N}(T)\ne \{0\}$, whereas a general partial isometry $V$ is not uniquely determined if $T$ has a non-trivial null space. $V$ is ...


1

This question has been addressed ad infinitum. For proof that $C^0$ is complete with the supremum norm (hence is a Banach space) see here: How to show that $C=C[0,1]$ is a Banach space For proof that the supremum norm arises from no inner product (and hence is not a Hilbert space) see here: $C[0,1]$ is not Hilbert space


1

The notation $|T|$ means the unique positive square root of $T^{\star}T$. Notice that $$ \||T|x\|^{2}=(|T|x,|T|x)=(|T|^{2}x,x)=(T^{\star}Tx,x)=(Tx,Tx)=\|Tx\|^{2}. $$ So $\mathcal{N}(T)=\mathcal{N}(|T|)$. Let $Y=\mathcal{R}(|T|)^{c}$. Then $|T| : Y\rightarrow Y$ is invertible because $\mathcal{N}(|T|)\cap ...


1

We have $T=V|T|$. Left multiply both sides by $V^{*}$ to get $V^{*}T=V^{*}V|T|$. Since $V$ is a partial isometry on the closure of the range of $|T|$, $V^{*}V=I$ on Im $|T|$ whence $V^{*}T=|T|$.


1

Let $\{ e_{j}\}_{j=1}^{N}$ be an orthonormal basis of $\mathcal{R}(T^{\star}T)$--which is finite-dimensional and, hence, closed--and extend this to a full orthonormal basis of $X$ by adding an orthonormal basis $\{ e_{j} \}_{j=N+1}^{\infty}$ of $\mathcal{R}(T^{\star}T)^{\perp}=\mathcal{N}(T^{\star}T)=\mathcal{N}(T)$. Then $Te_{j}=0$ for $j > N$, and $$ ...


1

I think I have answered this for you before, but I'll explain it again in more detail. The proof hinges on the following lemma, whose proof is given at the end. Perhaps you'll accept this answer. Lemma Let $X$ be a complex Hilbert space and let $T\in\mathcal{L}(X)$ satisfy $(Tx,x) \ge 0$ for all $x \in X$. Then $(Tx,x)=0$ iff $Tx=0$. That is, ...


1

I don't see the point of your last computation: you have already obtained what you are looking for before the "And if we take..." You got your $e_i$ and $f_i$ in the wrong spots, but that's just a labeling issue: you should have started with $\{f_i\}$ as the orthonormal basis. On a deeper side, why would $\{T^*e_i\}$ and orthonormal basis? In general, it ...


0

Hint: $$P^2(v)=P(P(v))=\frac{\langle P(v),w\rangle}{\|w\|^2}w$$


4

We have $$\require{cancel}P^2(v)=P(P(v))=P\left(\frac{\langle v,w\rangle }{||w||^2}w\right)=\frac{\langle v,w\rangle }{||w||^2}P\left(w\right)=\frac{\langle v,w\rangle }{||w||^2}\cancelto{=1}{\frac{\langle w,w\rangle }{||w||^2}}w=P(v)$$ Moreover, we have by the Cauchy-Schwarz inequality $$||P(v)||=\left|\frac{\langle v,w\rangle ...


2

As $A$ and $B$ commute, you have $\langle (A^2-B^2)x,x\rangle=0$ for all $x$, so $A^2=B^2$. Now you can use that both $A,B$ are positive, so $A$ is the unique positive square root of $A^2$, and likewise with $B$. So $A=B$. I don't immediately see how to prove that $A=B$ in the kernel of $A+B$ other than doing what I did above, which shows $A=B$ everywhere. ...


0

Actually, $D$ being closed has nothing to do with it @AndresCaicedo. $<Tx,x>=0$ for all $x$ in $D$ implies $T=0$ on $D$ precisely when $D$ is a subspace of $H$ and $T(D)$ is contained in $D$. This can be seen easily if you proof by the polarization identity. Then at the end of the proof you get $<Tx,y>=0$ for all $x,y$ in $D$. Then you need to be ...


0

Let $X$ be a complex Hilbert space. For any $T\in\mathcal{L}(X)$ (normal or not), there is a unique $P\in\mathcal{L}(X)$ such that $P \ge 0$ and $P^{2}=T^{\star}T$. So your question is confusing. If you define $|T|$ to be a positive square root of $T^{\star}T$, then any $P \ge 0$ for which $P^{2}=T^{\star}T$ must be $|T|$.


0

Note that Rudin requires the $U$ in the polar decomposition to be a unitary. In that case, if $T=UP$, then $$ T^*T=PU^*UP=P^2, $$ and indeed $P=|T|$. So there is no ambiguity.


1

To summarize the comments in an answer: You are correct, the topology on $I$ is irrelevant and was probably mentioned by mistake. (If someone will upvote this answer, the question will stop being bumped.)


0

The definition of $f(T)$ is $$ f(T) = \int_{\mathbb{R}} f(\lambda)dE(\lambda), $$ for any bounded Borel function $f$ on $\sigma(T)$. For convenience, extend $f$ to be $0$ on $\mathbb{C}\setminus\sigma(T)$, and suppose $f$ is strictly bounded by $M$. Then $\Re f$ and $\Im f$ are strictly bounded by $M$. Let $N$ be a positive integer, and divide the ...


0

The point is that, at the function level, you have $$ 1_w\circ f=1_{f^{-1}(w)}. $$ So, using functional calculus, $$ E_{f(T)}(w)=1_w(f(T))=1_w\circ f(T)=1_{f^{-1}(W)}(T)=E_T(f^{-1}(w)). $$



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